Numpy: placing values into an 1-of-n array based on indices in another array

Question

Suppose we had two arrays: some values, e.g. array([1.2, 1.4, 1.6]), and some indices (let's say, array([0, 2, 1])) Our output is expected to be the values put into a bigger array, "addressed" by the indices, so we would get

array([[ 1.2,  0. ,  0. ],
       [ 0. ,  0. ,  1.4],
       [ 0. ,  1.6,  0. ]])

Is there a way to do this without loops, in a nice, fast way?

JBernardo · Accepted Answer · 2012-10-30 03:12:29Z

2

With

a = zeros((3,3))
b = array([0, 2, 1])
vals = array([1.2, 1.4, 1.6])

You just need to index it (with the help of arange or r_):

>>> a[r_[:len(b)], b] = vals


array([[ 1.2,  0. ,  0. ],
       [ 0. ,  0. ,  1.4],
       [ 0. ,  1.6,  0. ]])

answered Oct 30, 2012 at 3:12

JBernardo

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1 Comment

Pierre GM Over a year ago

As an alternative: a[zip(*enumerate(b))]=vals

Kaje · Accepted Answer · 2013-11-29 03:49:34Z

0

How do we modify this for higher dimensions? For example, a is a 5x4x3 array and b and vals are 5x4 arrays. then How do we modify the statement a[r_[:len(b)],b] = vals ?

answered Nov 29, 2013 at 3:49

Kaje

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